Seven resistances are connected as shown in t | vedclass

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(B) $1$. First,simplify the parallel combination of two $10 \ \Omega$ resistors: $R_p = \frac{10 	imes 10}{10 + 10} = 5 \ \Omega$.$2$. Next,simplify

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$4$

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(B) $1$. First,simplify the parallel combination of two $10 \ \Omega$ resistors: $R_p = \frac{10 	imes 10}{10 + 10} = 5 \ \Omega$.$2$. Next,simplify the parallel combination of two $6 \ \Omega$ resistors: $R_p' = \frac{6 	imes 6}{6 + 6} = 3 \ \Omega$.$3$. The circuit now reduces to a Wheatstone bridge structure with arms $5 \ \Omega, 5 \ \Omega, 3 \ \Omega, 3 \ \Omega$ and a central $8 \ \Omega$ resistor.$4$. Since the ratio of resistances in the arms is equal $(\frac{5}{5} = \frac{3}{3} = 1)$,the bridge is balanced,and no current flows through the central $8 \ \Omega$ resistor.$5$. The circuit simplifies to two branches in parallel,each having a resistance of $(5 + 3) = 8 \ \Omega$.$6$. The equivalent resistance $R_{AB} = \frac{8 	imes 8}{8 + 8} = \frac{64}{16} = 4 \ \Omega$.

Seven resistances are connected as shown in the figure. The equivalent resistance between $A$ and $B$ is ............... $\Omega$.

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